设$m,n$是两个正整数且$m>n$,证明:$$\sum_{k=1}^{m-1} \sin^{2n} \frac{k\pi}{m}=\frac{mC_{2n}^n}{2^{2n}}$$
证明:
令$z=\exp(\frac{\pi}{m}i)$,则$|z|=1,z^{2m}=1$\[(z-\frac{1}{z})^{2n}=\sum_{k=0}^{2n} C_{2n}^k z^{2n-k}(-\frac{1}{z})^k=\sum_{k=0}^{2n}(-1)^k C_{2n}^k z^{2n-2k}=\sum_{j=-n}^{n}(-1)^{n-j} C_{2n}^{n-j} z^{2j}\]
接下来,用$z^r,r=0,1,2,\ldots,m-1$替换上式的$z$,并对所有可能值求和\begin{gather*}\sum_{r=0}^{m-1} (z^r-\frac{1}{z^r})^{2n}=\sum_{j=-n}^{n} (-1)^{n-j} C_{2n}^{n-j}(\sum_{r=0}^{m-1} z^{2rj})\end{gather*}
由于$|2j| \le 2n<2m$,故当$j \ne 0,z^{2j} \ne 1$,此时总有\[\sum_{r=0}^{m-1} z^{2rj}=\frac{z^{2mj}-1}{z^{2j}-1}=0\]
从而\[\sum_{r=0}^{m-1} (z^r-\frac{1}{z^r})^{2n}=(-1)^{n} C_{2n}^{n} \cdot m\]
又有\[\sum_{r=0}^{m-1} (z^r-\frac{1}{z^r})^{2n}=\sum_{r=0}^{m-1} (2i\sin\frac{r\pi}{m})^{2n}=(-1)^n \cdot 2^{2n} \sum_{r=0}^{m-1} \sin^{2n}\frac{r\pi}{m}\]
故\[(-1)^{n} C_{2n}^{n} \cdot m=(-1)^n \cdot 2^{2n} \sum_{r=0}^{m-1} \sin^{2n}\frac{r\pi}{m}\]
即\[\sum_{r=1}^{m-1} \sin^{2n}\frac{r\pi}{m}=\sum_{r=0}^{m-1} \sin^{2n}\frac{r\pi}{m}=\frac{m C_{2n}^n}{2^{2n}}\]
